Tyre Rotation for Maximum Distance
A car has two front tyres and two rear tyres. Front tyres wear out after $x$ km, and rear tyres wear out after $y$ km (with $x \neq y$ in general). You can swap the front and rear tyres at any point during the journey.
What is the maximum total distance the car can travel before any tyre is completely worn out? At what point should you perform the swap?
Hints
- Each tyre has a wear budget of 1. A tyre in the front position consumes $1/x$ per km; in the rear position it consumes $1/y$ per km. The goal is to exhaust all tyres simultaneously.
- Let $s$ be the swap point and $d$ the total distance. Write the wear equation for a single tyre: $s/x + (d-s)/y = 1$. By symmetry the same equation applies to all four tyres.
- Add the wear equations for both tyre types to eliminate $s$. You will get $d(1/x + 1/y) = 2$, which gives $d = 2xy/(x+y)$ directly.
Worked Solution
How to Think About It: Each tyre has a wear budget. A front tyre consumes $1/x$ of its life per km, and a rear tyre consumes $1/y$ per km. If you swap at some point $s$, each tyre spends some distance in front and some in rear. The goal is to choose $s$ so that all four tyres reach exactly 100% wear at the same time -- that way no tyre goes to waste and no tyre blows out early.
Quick Estimate: Say $x = 40{,}000$ km (front) and $y = 80{,}000$ km (rear). Harmonic mean: $2 \times 40{,}000 \times 80{,}000 / (40{,}000 + 80{,}000) = 6.4 \times 10^9 / 120{,}000 \approx 53{,}333$ km. So the answer is between the two limits ($40k$ and $80k$), closer to the smaller one. Swap at $53{,}333/2 \approx 26{,}667$ km.
Approach: Set up a wear equation per tyre and solve for the total distance $d$ that exhausts all tyres simultaneously.
Formal Solution:
Let $d$ be the total distance and $s$ be the distance at which we swap. Each tyre spends $s$ km in one position and $d - s$ km in the other.
For a tyre that starts in front and moves to rear, its total wear is: $$\frac{s}{x} + \frac{d - s}{y} = 1$$
For a tyre that starts in rear and moves to front: $$\frac{s}{y} + \frac{d - s}{x} = 1$$
By symmetry, both equations give the same constraint. Adding them: $$\frac{s}{x} + \frac{d-s}{y} + \frac{s}{y} + \frac{d-s}{x} = 2$$ $$d\left(\frac{1}{x} + \frac{1}{y}\right) = 2$$ $$d = \frac{2xy}{x + y}$$
The optimal swap point is $s = d/2 = \frac{xy}{x+y}$ km.
Answer: $$d = \frac{2xy}{x + y}$$ Swap at $s = \frac{xy}{x + y}$ km. The maximum distance is the harmonic mean of $x$ and $y$.
Intuition
The answer is the harmonic mean of $x$ and $y$, which always lies between the two values and is closer to the smaller one. This makes sense: if front tyres wear faster, the harmonic mean is pulled toward $x$ -- you cannot escape the bottleneck of the faster-wearing tyre entirely, but swapping lets you share the burden.
The general principle here is that when you have two resources consumed at different rates and you can switch between them, the optimal strategy is to equalize total consumption across all units. In this case that means every tyre does exactly half its life in front and half in rear. This equal-sharing insight recurs across quant problems -- from load balancing in distributed systems to risk budgeting across correlated assets.